Question

7．代數(shù)式ad-bc可用符號(hào)$|\begin{array}{l}a{\;}^{\;}{\;}_{\;}b\\ c{\;}^{\;}{\;}_{\;}d\end{array}|$來(lái)表示，稱(chēng)之為二階行列式．即$|\begin{array}{l}a{\;}^{\;}{\;}_{\;}b\\ c{\;}^{\;}{\;}_{\;}d\end{array}|=ad-bc$，用二階行列式可以解二元一次方程組．由$\left\{\begin{array}{l}{a_1}x+{b_1}y={c_1}\\{a_2}x+{b_2}y={c_2}\end{array}\right.$得三個(gè)二階行列式即$D=|\begin{array}{l}{a_1}{\;}^{\;}{b_1}\\{a_2}{\;}^{\;}{b_2}\end{array}|$，${D_x}=|\begin{array}{l}{c_1}{\;}^{\;}{b_1}\\{c_2}{\;}^{\;}{b_2}\end{array}|$及${D_y}=|\begin{array}{l}{a_1}{\;}^{\;}{c_1}\\{a_2}{\;}^{\;}{c_2}\end{array}|$那么方程組的解就是$\left\{\begin{array}{l}x=\frac{D_x}{D}\\ y=\frac{D_y}{D}\end{array}\right.$．
（1）求出二階行列式$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$的值；
（2）用二階行列式解方程組$\left\{\begin{array}{l}3x+2y=-1\\ 5x-y-2=0\end{array}\right.$．

Answer 1

分析 （1）根據(jù)題意可以直接算出二階行列式$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$的值；
（2）根據(jù)題意可以算出D、D_X，D_Y，從而可以求得x、y的值，本題得以解決．

解答 解：（1）由題意可得，
$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$=3×4-5×6=12-30=-18，
即$|\begin{array}{l}3{\;}^{\;}{\;}_{\;}5\\ 6{\;}^{\;}{\;}_{\;}4\end{array}|$的值是-18；
（2）∵$\left\{\begin{array}{l}{3x+2y=-1}\\{5x-y-2=0}\end{array}\right.$，
∴$\left\{\begin{array}{l}{3x+2y=-1}\\{5x-y=2}\end{array}\right.$，
由題意可得，
D=$|\begin{array}{l}{3}&{2}\\{5}&{-1}\end{array}|$=3×（-1）-2×5=-3-10=-13，
${D}_{X}=|\begin{array}{l}{-1}&{2}\\{2}&{-1}\end{array}|$=（-1）×（-1）-2×2=1-4=-3，
${D}_{Y}=|\begin{array}{l}{3}&{-1}\\{5}&{2}\end{array}|$=3×2-（-1）×5=6+5=11，
∴$x=\frac{{D}_{X}}{D}=\frac{-3}{-13}=\frac{3}{13}$，
$y=\frac{{D}_{Y}}{D}=\frac{11}{-13}=-\frac{11}{13}$，
方程組的解是$\left\{\begin{array}{l}{x=\frac{3}{13}}\\{y=-\frac{11}{13}}\end{array}\right.$．

點(diǎn)評(píng) 本題考查二元一次方程組的解和解二元一次方程，解題的關(guān)鍵是明確題目中的新定義，根據(jù)新定義可以解決相關(guān)的問(wèn)題．